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Intern  Joined: 30 May 2008
Posts: 45
From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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204 00:00

Difficulty:   65% (hard)

Question Stats: 51% (01:12) correct 49% (01:11) wrong based on 3590 sessions

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From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A. (-10)^20
B. (-10)^10
C. 0
D. –(10)^19
E. –(10)^20
Math Expert V
Joined: 02 Sep 2009
Posts: 64949
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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catty2004 wrote:
From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A. (-10)^20
B. (-10)^10
C. 0
D. –(10)^19
E. –(10)^20

Select 10 odd number of times and -10 the remaining number of times, for example select 10 once and -10 nineteen times, then the product will be $$10^1*(-10)^{19}=-10^{20}$$.

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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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1
7
The minimum value is the biggest whole number with a negative sign.

$$(-10)*(10)^{19}$$

But this is not anywhere in the answer choice, let's manipulate: $$(-10)(10)^{20}=(-1)(10)(10)^{19}=-10^{20}$$

##### General Discussion
Intern  Joined: 30 May 2008
Posts: 45
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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Bunuel wrote:
catty2004 wrote:
From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A. (-10)^20
B. (-10)^10
C. 0
D. –(10)^19
E. –(10)^20

Select 10 odd number of times and -10 the remaining number of times, for example select 10 once and -10 nineteen times, then the product will be $$10^1*(-10)^{19}=-10^{20}$$ .

I'm not quiet sure i understand, why choose only -10 and 10, there are other integers included in the set of -10 to 10. Also, isn't 10^1*10^(-19) = 10^(-18)?
Math Expert V
Joined: 02 Sep 2009
Posts: 64949
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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1
catty2004 wrote:
Bunuel wrote:
catty2004 wrote:
From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A. (-10)^20
B. (-10)^10
C. 0
D. –(10)^19
E. –(10)^20

Select 10 odd number of times and -10 the remaining number of times, for example select 10 once and -10 nineteen times, then the product will be $$10^1*(-10)^{19}=-10^{20}$$ .

I'm not quiet sure i understand, why choose only -10 and 10, there are other integers included in the set of -10 to 10. Also, isn't 10^1*10^(-19) = 10^(-18)?

We choose -10 and 10 because this way we get the least product (notice that -10^(20) is the smallest number among answer choices).

Also it's $$10^1*(-10)^{19}=-10^{20}$$ not $$10^1*(10)^{-19}=-10^{20}$$.
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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Thanks for pointing that out!!! Silly me. But i'm still not sure, why the other answer choices are eliminated and why E is the least possible answer?? >_<
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Joined: 16 Oct 2010
Posts: 10629
Location: Pune, India
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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5
catty2004 wrote:
From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A. (-10)^20
B. (-10)^10
C. 0
D. –(10)^19
E. –(10)^20

The consecutive integers from -10 to 10 are: -10, -9, -8, -7 ... -1, 0, 1, ... 9, 10

You have to select 20 integers with repetition.

So you can select all 10. The product will be $$10^{20}$$.
or
You can select all -10. The product will be $$(-10)^{20}$$ (which is essentially same as $$10^{20}$$)
or
You can select ten 1s and ten -10s. The product will be $$(-10)^{10}$$
or
You can select 0 and any other 19 numbers. The product will be 0.
or
You can select -1 and nineteen 10s. The product will be $$-(10)^{19}$$.

But how will you get $$- (10)^{20}$$?

Editing here itself - I missed out the case in which you can get $$- (10)^{20}$$ (by multiplying odd number of -10 with rest of the 10s). Possibly because I mis-read the question.
'Least possible value of the product' to me implied the product that is not possible. Since all options a to d were possible, I overlooked that option (E) is possible too.

Actually the question should read 'the smallest possible value of the product' to avoid such confusion. Since -10^20 is the smallest value that can be obtained (greatest magnitude but negative), the answer is (E).
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Karishma
Veritas Prep GMAT Instructor

Originally posted by VeritasKarishma on 08 Jul 2012, 19:59.
Last edited by VeritasKarishma on 09 Jul 2012, 02:00, edited 1 time in total.
Math Expert V
Joined: 02 Sep 2009
Posts: 64949
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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2
VeritasPrepKarishma wrote:
catty2004 wrote:
From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A. (-10)^20
B. (-10)^10
C. 0
D. –(10)^19
E. –(10)^20

The consecutive integers from -10 to 10 are: -10, -9, -8, -7 ... -1, 0, 1, ... 9, 10

You have to select 20 integers with repetition.

So you can select all 10. The product will be $$10^{20}$$.
or
You can select all -10. The product will be $$(-10)^{20}$$ (which is essentially same as $$10^{20}$$)
or
You can select ten 1s and ten -10s. The product will be $$(-10)^{10}$$
or
You can select 0 and any other 19 numbers. The product will be 0.
or
You can select -1 and nineteen 10s. The product will be $$-(10)^{19}$$.

But how will you get $$- (10)^{20}$$?
You need to select -1 and twenty 10s but you cannot select 21 numbers. You cannot have the product as negative $$10^{20}$$.
Hence (E) is not possible
.

E is possible if you select 10 odd number of times and -10 the remaining number of times. For example:

$$10^1*(-10)^{19}=-10^{20}$$ or $$10^3*(-10)^{17}=-10^{20}$$ or $$10^5*(-10)^{15}=-10^{20}$$... or $$10^{19}*(-10)^{1}=-10^{20}$$.

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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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catty2004 wrote:
But i'm still not sure, why the other answer choices are eliminated and why E is the least possible answer?? >_<

Actually, the question means "smallest possible value of the product"

$$-10^{20}$$ is the smallest value of the given 5 options. Since it is possible, so it is the answer. I guess you were thinking 'least possible' (as was I). The wording is definitely misleading.
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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VeritasPrepKarishma wrote:
catty2004 wrote:
But i'm still not sure, why the other answer choices are eliminated and why E is the least possible answer?? >_<

Actually, the question means "smallest possible value of the product"

$$-10^{20}$$ is the smallest value of the given 5 options. Since it is possible, so it is the answer. I guess you were thinking 'least possible' (as was I). The wording is definitely misleading.

Yes, the question means "what is the smallest possible value of the product of the 20 integers?"
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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3
i missed out the bracket for -10^20. I was assuming that value to be positive as a negative number raised to positive power is positive. Nice problem![/edit]
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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Bigred2008 wrote:
From the consecutive integers -10 to 10, inclusive 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of 20 integers.

A. (-10)^20
B. (-10)^10
C. 0
D. -(10)^19
E. -(10)^20

The smallest possible value of the product will be negative (since it has to be smallest) with highest absolute value i.e. the number should be as left on the number line as possible.
Out of all the options, (E) is the smallest. Let's see if it is possible.
If you select all 10s and only one -10, you will get $$-10*(10)^{19}$$ which is the same as $$-(10)^{20}$$. So it is possible to get the product given in (E).

Notice that options (A) and (B) are positive.
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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This question shows how important it is to carefully look at the symbols. One might be very easily deceived in thanking the question is actually saying --> (-10)^20.
Intern  Joined: 29 Aug 2013
Posts: 11
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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Hi guys,

just to make sure, picking "-10" 19x and "+10" the remaining time would also work, right?

-> (-10)^19 * 10^1

?

Many thanks
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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mott wrote:
Hi guys,

just to make sure, picking "-10" 19x and "+10" the remaining time would also work, right?

-> (-10)^19 * 10^1

?

Many thanks

Yes, it does work. That is how you get $$-(10)^{20}$$.
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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1
Hi,

I also chose the right answer quite quickly. However, I didn't use powers and I am not sure why we used powers.

So what I though it that we need an odd number of negative values to have a negative value.
To have the greatest negative value we need -10 (the higest possible value), the greatest number of odd times; so in out case 19.
We still have to use one number, so 10 would be the final choise, as it would make the negative 19(-10) 10 times greater.

However, for me it is:
-10(19) * 10 (1) instead of 10^1 * -10^19 = -10^20.

I understand the 10^1, but -10(19) is not -10^19. This is where I am lost...

So, why the powers?
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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pacifist85 wrote:
Hi,

I also chose the right answer quite quickly. However, I didn't use powers and I am not sure why we used powers.

So what I though it that we need an odd number of negative values to have a negative value.
To have the greatest negative value we need -10 (the higest possible value), the greatest number of odd times; so in out case 19.
We still have to use one number, so 10 would be the final choise, as it would make the negative 19(-10) 10 times greater.

However, for me it is:
-10(19) * 10 (1) instead of 10^1 * -10^19 = -10^20.

I understand the 10^1, but -10(19) is not -10^19. This is where I am lost...

So, why the powers?

You are given that you need to find the product of all 20 numbers. When you multiply -10 by -10, you get $$(-10)^2$$, not $$-10*2$$ (which is -10 + (-10))

Just like you multiplied the last 10 you picked, you have to multiply all previous 10s too.
So multiply -10, 19 times to get $$-10*-10*-10...*-10 = (-10)^{19}$$
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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mbaspire wrote:
From the consecutive integers -10 to 10, inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A $$(-10)^{20}$$
B $$(-10)^{10}$$
C 0
D $$-(10)^{19}$$
E $$-(10)^{20}$$

Hi,

we have to choose all 10s and ensure the combination is such that the sign becomes -ive..
$$10^{19}*(-10)..$$
or $$(-10)^{19}*10= -10^{20}$$

E
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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2
mbaspire wrote:
From the consecutive integers -10 to 10, inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A $$(-10)^{20}$$
B $$(-10)^{10}$$
C 0
D $$-(10)^{19}$$
E $$-(10)^{20}$$

Choose nineteen -10's and one 10
So, the product = [(-10)^19]
Notice that [(-10)^19] is NEGATIVE, which means [(-10)^19] is also NEGATIVE.

So, [(-10)^19] = -[(10)^19]
= -(10)^20
= E

Cheers,
Brent
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Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

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Bunuel wrote:
mbaspire wrote:
From the consecutive integers -10 to 10, inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

A $$(-10)^{20}$$
B $$(-10)^{10}$$
C 0
D $$-(10)^{19}$$
E $$-(10)^{20}$$

Hi Bunuel,

I'm confused. I have referred to gmat club math book about multiply 2 number with exponents. the base should be the same to allow sum of the exponent while in this question the base is 10 & -10. So we can't sum 10^10 * (-10)^19. Do I miss something?

Thanks Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran   [#permalink] 28 Mar 2016, 00:42

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